opplettmcgx

2023-03-25

What are the coordinates of the center and the length of the radius of the circle represented by the equation ${x}^{2}+{y}^{2}-4x+8y+11=0$?

Fleetuictr0th

Beginner2023-03-26Added 5 answers

The equation must be transformed to in order to determine the radius and center.

$(x-a)}^{2}+{(y=b)}^{2}={r}^{2$

${x}^{2}+{y}^{2}-4x+8y+11=0$

${x}^{2}-4x+4-4+{y}^{2}+8y+16-16+11=0$

${(x-2)}^{2}-4+{(y+4)}^{2}-16+11=0$

${(x-2)}^{2}+{(y+4)}^{2}-9=0$

${(x-2)}^{2}+{(y+4)}^{2}=9$

According to the transformed equation, the circle's center is C=(2;-4) and its radius is r=3.

$(x-a)}^{2}+{(y=b)}^{2}={r}^{2$

${x}^{2}+{y}^{2}-4x+8y+11=0$

${x}^{2}-4x+4-4+{y}^{2}+8y+16-16+11=0$

${(x-2)}^{2}-4+{(y+4)}^{2}-16+11=0$

${(x-2)}^{2}+{(y+4)}^{2}-9=0$

${(x-2)}^{2}+{(y+4)}^{2}=9$

According to the transformed equation, the circle's center is C=(2;-4) and its radius is r=3.

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